76.27.15 problem 15

Internal problem ID [17790]
Book : Differential equations. An introduction to modern methods and applications. James Brannan, William E. Boyce. Third edition. Wiley 2015
Section : Chapter 6. Systems of First Order Linear Equations. Section 6.5 (Fundamental Matrices and the Exponential of a Matrix). Problems at page 430
Problem number : 15
Date solved : Monday, March 31, 2025 at 04:27:46 PM
CAS classification : system_of_ODEs

\begin{align*} \frac {d}{d t}x_{1} \left (t \right )&=-3 x_{1} \left (t \right )-9 x_{2} \left (t \right )\\ \frac {d}{d t}x_{2} \left (t \right )&=x_{1} \left (t \right )-3 x_{2} \left (t \right ) \end{align*}

With initial conditions

\begin{align*} x_{1} \left (0\right ) = 4\\ x_{2} \left (0\right ) = 1 \end{align*}

✓ Maple. Time used: 0.142 (sec). Leaf size: 44

ode:=[diff(x__1(t),t) = -3*x__1(t)-9*x__2(t), diff(x__2(t),t) = x__1(t)-3*x__2(t)]; 
ic:=x__1(0) = 4x__2(0) = 1; 
dsolve([ode,ic]);
 

\begin{align*} x_{1} \left (t \right ) &= {\mathrm e}^{-3 t} \left (4 \cos \left (3 t \right )-3 \sin \left (3 t \right )\right ) \\ x_{2} \left (t \right ) &= -\frac {{\mathrm e}^{-3 t} \left (-3 \cos \left (3 t \right )-4 \sin \left (3 t \right )\right )}{3} \\ \end{align*}

✓ Mathematica. Time used: 0.006 (sec). Leaf size: 49

ode={D[x1[t],t]==-3*x1[t]-9*x2[t],D[x2[t],t]==1*x1[t]-3*x2[t]}; 
ic={x1[0]==4,x2[0]==1}; 
DSolve[{ode,ic},{x1[t],x2[t]},t,IncludeSingularSolutions->True]
 

\begin{align*} \text {x1}(t)\to e^{-3 t} (4 \cos (3 t)-3 \sin (3 t)) \\ \text {x2}(t)\to \frac {1}{3} e^{-3 t} (4 \sin (3 t)+3 \cos (3 t)) \\ \end{align*}

✓ Sympy. Time used: 0.104 (sec). Leaf size: 56

from sympy import * 
t = symbols("t") 
x__1 = Function("x__1") 
x__2 = Function("x__2") 
ode=[Eq(3*x__1(t) + 9*x__2(t) + Derivative(x__1(t), t),0),Eq(-x__1(t) + 3*x__2(t) + Derivative(x__2(t), t),0)] 
ics = {} 
dsolve(ode,func=[x__1(t),x__2(t)],ics=ics)
 

\[ \left [ x^{1}{\left (t \right )} = - 3 C_{1} e^{- 3 t} \sin {\left (3 t \right )} - 3 C_{2} e^{- 3 t} \cos {\left (3 t \right )}, \ x^{2}{\left (t \right )} = C_{1} e^{- 3 t} \cos {\left (3 t \right )} - C_{2} e^{- 3 t} \sin {\left (3 t \right )}\right ] \]